To apply the Mean Value Theorem (MVT) to the function $y(x) = x^3$ on the interval $[ -7, 9 ]$, we need to follow these steps:

Step 1: Conditions for the MVT

For the Mean Value Theorem to apply, two conditions must be satisfied:

1. The function must be continuous on the closed interval $[a, b]$.
2. The function must be differentiable on the open interval $(a, b)$.

The function $y(x) = x^3$ is both continuous and differentiable everywhere, so the MVT applies.

Step 2: Mean Value Theorem Statement

The Mean Value Theorem states that there exists at least one point $c$ in the open interval $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In our case, $f(x) = x^3$, $a = -7$, and $b = 9$.

Step 3: Calculate $f(b)$ and $f(a)$

$f(9) = 9^3 = 729$ $f(-7) = (-7)^3 = -343$

Now, compute the difference quotient:

$\frac{f(9) - f(-7)}{9 - (-7)} = \frac{729 - (-343)}{9 + 7} = \frac{729 + 343}{16} = \frac{1072}{16} = 67$

Step 4: Find $f'(x)$

The derivative of $f(x) = x^3$ is:

$f'(x) = 3x^2$

Now, we need to find $c$ such that:

$f'(c) = 67$

Substitute $f'(c) = 3c^2$:

$3c^2 = 67$

Solve for $c^2$:

$c^2 = \frac{67}{3} \approx 22.33$

Take the square root of both sides:

$c = \pm \sqrt{\frac{67}{3}} \approx \pm 4.73$

Step 5: Conclusion

The points $c$ that satisfy the conclusion of the MVT are approximately $c = 4.73$ and $c = -4.73$, which both lie within the interval $(-7, 9)$.

Would you like further details or explanations on any of these steps?

Here are 5 related questions:

1. How do you know if a function is differentiable on an interval?
2. What would happen if the function were not differentiable at any point in the interval?
3. How would the process change if the function were a polynomial of degree higher than 3?
4. Can the Mean Value Theorem be applied to a function with a vertical asymptote in the interval?
5. What is the geometric interpretation of the Mean Value Theorem?

Tip: Always check the conditions of continuity and differentiability before applying the MVT to a function.